What is the integral of (tan(x))/(1+cos^4(x)) ?

The integral of (tan(x))/(1+cos^4(x)) is -1/4 (ln(cos^4(x))-ln|1+cos^4(x)|)+C

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integral of (tan(x))/(1+cos^4(x))

\int \frac{tan ( x )}{1 + cos ^{4} ( x )} d x

- \frac{1}{4} (ln (cos^{4} (x)) - ln 1 + cos^{4} (x)) + C

Solution steps

\int \frac{tan ( x )}{1 + cos ^{4} ( x )} d x

Apply u-substitution

: \int - \frac{1}{u ( u ^{4} + 1 )} d u

= \int - \frac{1}{u ( u ^{4} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{u ( u ^{4} + 1 )} d u

Apply u-substitution

: \int \frac{1}{4 ( v - 1 ) v} d v

= - \int \frac{1}{4 ( v - 1 ) v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{4} \cdot \int \frac{1}{( v - 1 ) v} d v

Take the partial fraction of

\frac{1}{( v - 1 ) v} : \frac{1}{v - 1} - \frac{1}{v}

= - \frac{1}{4} \cdot \int \frac{1}{v - 1} - \frac{1}{v} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \frac{1}{4} (\int \frac{1}{v - 1} d v - \int \frac{1}{v} d v)

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

\int \frac{1}{v} d v = ln ∣ v ∣

= - \frac{1}{4} (ln ∣ v - 1 ∣ - ln ∣ v ∣)

Substitute back

= - \frac{1}{4} (ln 1 + cos^{4} (x) - 1 - ln 1 + cos^{4} (x))

Simplify

- \frac{1}{4} (ln 1 + cos^{4} (x) - 1 - ln 1 + cos^{4} (x)) : - \frac{1}{4} (ln (cos^{4} (x)) - ln 1 + cos^{4} (x))

= - \frac{1}{4} (ln (cos^{4} (x)) - ln 1 + cos^{4} (x))

Add a constant to the solution

= - \frac{1}{4} (ln (cos^{4} (x)) - ln 1 + cos^{4} (x)) + C

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What is the integral of (tan(x))/(1+cos^4(x)) ?
The integral of (tan(x))/(1+cos^4(x)) is -1/4 (ln(cos^4(x))-ln|1+cos^4(x)|)+C